Complete the square to rewrite this expression in the form $(x + a)^2 + b$. $a$ and $b$ can be positive or negative. $ x^{2}+2x+78$
Explanation: If we square the sum of $x$ and another number, we get $ \begin{align} (x + a)^2 &= (x + a)(x + a) \\ &= x^2 + {2a}x + a^2\end{align}$ The number multiplying the $x$ term is $2$ times the number that was added inside the square. In the problem we're trying to solve, the number multiplying the $x$ term is $2$ , so when we rewrite the expression, the number added inside the square will be half of $2$ . (Which is $1$ $ (x + 1)^2 + b$ How can we find the value of $b$ If we multiply out the square in this expression, we get $ \begin{align} (x + 1)^2 + b &= (x + 1)(x + 1) + b \\ &= x^2 + 2x + 1 + b \end{align}$ This looks just like the given expression if $ 1 + b = 78$ So $b$ must be $77$ $x^{2}+2x+78$ can be rewritten as: $ (x + 1)^2 + 77$ Now it's just a square minus another number! We have completed the square.